# 19.1 Electric potential energy: potential difference  (Page 2/10)

 Page 2 / 10
$V=\frac{\text{PE}}{q}\text{.}$

## Electric potential

This is the electric potential energy per unit charge.

$V=\frac{\text{PE}}{q}$

Since PE is proportional to $q$ , the dependence on $q$ cancels. Thus $V$ does not depend on $q$ . The change in potential energy $\text{ΔPE}$ is crucial, and so we are concerned with the difference in potential or potential difference $\Delta V$ between two points, where

$\Delta V={V}_{\text{B}}-{V}_{\text{A}}=\frac{\Delta \text{PE}}{q}\text{.}$

The potential difference between points A and B, ${V}_{\text{B}}\phantom{\rule{0.25em}{0ex}}–\phantom{\rule{0.25em}{0ex}}{V}_{\text{A}}$ , is thus defined to be the change in potential energy of a charge $q$ moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

$\text{1 V = 1}\phantom{\rule{0.25em}{0ex}}\frac{J}{C}$

## Potential difference

The potential difference between points A and B, ${V}_{B}-{V}_{A}$ , is defined to be the change in potential energy of a charge $q$ moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

$\text{1 V = 1}\phantom{\rule{0.25em}{0ex}}\frac{J}{C}$

The familiar term voltage is the common name for potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor.

In summary, the relationship between potential difference (or voltage) and electrical potential energy is given by

$\Delta V=\frac{\text{ΔPE}}{q}\phantom{\rule{0.25em}{0ex}}\text{and ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V\text{.}$

## Potential difference and electrical potential energy

The relationship between potential difference (or voltage) and electrical potential energy is given by

$\Delta V=\frac{\text{ΔPE}}{q}\phantom{\rule{0.25em}{0ex}}\text{and ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V\text{.}$

The second equation is equivalent to the first.

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other since $\text{ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V$ . The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

## Calculating energy

Suppose you have a 12.0 V motorcycle battery that can move 5000 C of charge, and a 12.0 V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

Strategy

To say we have a 12.0 V battery means that its terminals have a 12.0 V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to $\text{ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V$ .

So to find the energy output, we multiply the charge moved by the potential difference.

Solution

For the motorcycle battery, $q=\text{5000 C}$ and $\Delta V=\text{12.0 V}$ . The total energy delivered by the motorcycle battery is

$\begin{array}{lll}{\text{ΔPE}}_{\text{cycle}}& =& \left(\text{5000 C}\right)\left(\text{12.0 V}\right)\\ & =& \left(\text{5000 C}\right)\left(\text{12.0 J/C}\right)\\ & =& 6.00×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}$

Similarly, for the car battery, $q=\text{60},\text{000}\phantom{\rule{0.25em}{0ex}}\text{C}$ and

$\begin{array}{lll}{\text{ΔPE}}_{\text{car}}& =& \left(\text{60,000 C}\right)\left(\text{12.0 V}\right)\\ & =& 7.20×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}$

Discussion

While voltage and energy are related, they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a low car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.

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